Question

A straight line moves so that the product of the perpendiculars on it from two fixed points is constant. Prove that the locus of the feet of the perpendiculars from each of these points upon the straight line is a circle, the same for each.

Answer 1

Let $A$ and $B$ be the two fixed points.

Taking axis of $X$ along $AB$ and the right bisector of $AB$ as $y$-axis, $A(-a,0)\&B(a,0)$, then
Let the equation of the given line be
$x\cos \alpha +y\sin \alpha =p....1$
Equation of the line through $(-a,0)$ and perpendicular to $1$ is
$y-0=\frac{\sin \alpha }{\cos \alpha }(x+a)$
$y\cos \alpha -x\sin \alpha =a\sin \alpha ....2$
$AN\times BM=\frac{-a\cos \alpha +0\sin \alpha -p}{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}\times \frac{a\cos \alpha +0\sin \alpha -p}{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}$
$={\lambda }^2$
or $p^2-a^2{\cos }^2\alpha ={\lambda }^2$
or $p^2={\lambda }^2+a^2{\cos }^2\alpha .....3$
Eliminate p and $\alpha$ from $1$ , $2$ and $3$; squaring $1$ and $2$ and then adding
$x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\cos \alpha \sin \alpha +y^2{\cos }^2\alpha +x^2{\sin }^2\alpha -2xy\cos \alpha \sin \alpha =p^2+a^2{\sin }^2\alpha$
or $x^2+y^2=p^2+a^2{\sin }^2\alpha$
Putting from $p^2$ from $3$
$x^2+y^2={\lambda }^2+a^2$

Locus of both the points since on changing $a$ to $-a$, there is no change in the equation.